for (i = 0; string[i] != '\0' && string[i] != '=';)
{
//从输入串中取出一个字符作为开始，进行处理，直到表达式结束
if (string[i] >= '0' && string[i] <= '9')
{
//如果是操作数，将整个操作数提取出来，压入操作数栈
pushNumber(getNumFromString(&string[i], &i));
}
else
{
op.name = string[i];
op.priority = getPriority(string[i]);
op.opnum = getOpNum(string[i]);
topOp = popOperator();
if (op.name == '(')
{
//如果是'('，将从栈顶弹出的运算符压回栈内，并将当前运算符则压栈
pushOperator(topOp);
pushOperator(op);
}
else if (op.name == ')')
{
//如果是')'，则进行运算，每次运算结果作为一个操作数压入操作数栈，直到将'('弹出运算符栈
while (topOp.name != '(')
{
pushNumber(operate(topOp));
topOp = popOperator();
}
}
else
{
//如果是普通运算符
if (topOp.name != '#' && op.priority <= topOp.priority)
{
//如果运算符栈非空，且当前运算符的优先级大于栈顶运算符，则进行一次运算，将结果压入操作数栈
pushNumber(operate(topOp));
}
else
{
//否则将从栈顶弹出的运算符压回
pushOperator(topOp);
}
//将当前运算符压栈
pushOperator(op);
}
i++;
}
}
//完成栈内剩余的运算
while ((topOp = popOperator()).name != '#')
{
pushNumber(operate(topOp));
}
//操作数栈中剩下的最后一个数即为结果
printf("%f\n", popNumber());
return 0;
}
作者：卞昊穹
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